Figure 2-3: (a) Rheostat is used to set the resistance between terminals 1 and 2 at any value between zero and Rmax; (b) location of wiper in

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3 Figure 2-3: (a) Rheostat is used to set the resistance between terminals 1 and 2 at any value between zero and Rmax; (b) location of wiper in potentiometer divides the resistance Rmax among R13 and R23.

4 Circuit Elements: Independent Sources

5 Circuit Elements: Dependent Sources

6 Kirchoff s laws: A node is a point where two are more circuit elements meet. Kirchoff s Current Law (KCL): The algebraic sum of all the currents at any node in a circuit equals zero. A closed-path or loop: Starting at any arbitrary node, walk along an element connected to it to move to another node, repeat this as often as necessary, and come back to the original node. The path you traced is a closed path or loop. Kirchoff s Voltage Law (KVL): The algebraic sum of all the voltages around any closed-path in a circuit equals zero.

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8 Two important types of interconnection of circuit components: Series Connection: Two or more elements (branches, components) are said to be interconnected in series if the currents through each one of them are equal. Parallel Connection: Two or more elements (branches, components) are said to be interconnected in parallel if the voltages across each one of them are equal. In the following diagrams, two elements are interconnected forming an equivalent one element with two terminals a and b. Decide as far as the terminals a and b are concerned, whether the two elements are interconnected in series or in parallel.

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11 Principles of Electrical Engineering I Interconnection of Sources v 1 v 2 i v v 1 v 2 i v Voltage sources connected in series Voltage sources connected in parallel i 1 i 2 i v i i 1 i 2 v Current sources connected in series Current sources connected in parallel All interconnections that do not satisfy Kirchoff s laws are illegal. Illustration of illegal interconnections: 3A 12V 9V 4A 3Ω

12 We often use two phrases `open circuit' between two terminals or `short circuit' between two terminals. In the following figure, terminals 1 and 2 are open circuited, while terminals 3 and 4 are short circuited. The terminology applies to only those specified terminals. A voltmeter reads the voltage across an element, as such it is connected across (or parallel) to the element. An ammeter reads the current through an element, as such it is connected in series to the element.

13 Basics you should always remember Electric Potential by itself is not power. Current has to be supplied or consumed to generate power. In other words, the interaction of Electric Potential and Electric current causes power to be generated or consumed. In the same way, your inherent potential by itself is not enough, it needs to be put to work by doing Home-Work in order to learn the material properly. Then only circuit analysis in its full detail can be absorbed by you. Voltage is across a branch and current is through a branch: i v Power generated by the small box = vi i v Power absorbed by the small box = vi Figure 1 Resistance: Water flows down the hill unless it is being pumped up. Current flows down the potential hill through a resistance. Resistance is a passive element. It always consumes power. Figure 2 Source is an active element. A source could pump the current up the potential hill and thus generate power. A source could have a current flowing down the potential hill and thus consume power; in this case it is a sink. Think of a good battery charging a weak one. In the figure given on the right, two batteries a good one (denoted by α) and a weak one (denoted by β) are interconnected at terminals a and b. Which of the two batteries generates power and which one consumes? Is i positive if v is positive? i R i i v = Ri R v = Ri R v = Ri I Z V = ZI, Z is the impedance α i b - v a β v i β For a voltage source, v is prescribed, and i depends on the circuit β. v i β For a current source, i is prescribed, and v depends on the circuit β. Voltage Source Current Source

14 In circuit analysis, different situations arise. Both the voltage and the current of a branch might be known ahead. In this case the branch info is completely known. Either voltage or current of a branch but not both are known. Neither voltage nor current of a branch are known. In the case of a voltage source, voltage is known ahead but not current. That is, the voltage of a voltage source is prescribed ahead but the current through it is a variable that depends on the network to which it is connected. In the case of a current source, current is known ahead but not voltage. That is, the current of a current source is prescribed ahead but the voltage across it is a variable that depends on the network to which it is connected. 2A i = 6A 12V 2Ω i = 2A 2Ω 4V i = 4A 12V 3Ω 2A i = 2A 4Ω 8V i = 0 12V Open Circuit 2A i = 2A In the case of a resistance, the ratio between voltage and current (via Ohms s law) is known. If a voltage or current is unknown, you can assign a variable to it. If we do so for each branch in the circuit, we end up with some unknown variables. The aim of circuit analysis is to write as many equations (via KCL, KVL, Ohms s law) as there are unknown variables, and then solve those equations to determine the unknown variables. The entire circuit analysis involves only this type of work. All we do in this course is to develop ways of simplifying the method(s) of writing equations, as small in number as needed. 0 V - Short Circuit Nodal Analysis: Consider a resistance R with two terminals marked A and B. Suppose the voltage of terminals A and B with respect to the ground G are respectively v a and v b as shown. Then the voltage across R is v = v a v b, and the current through R is i = vav b R. v a A i R v v b B G Mesh Analysis: Consider a resistance R having two currents i 1 and i 2 pumped into it as shown. Then the net current through it is i = i 1 i 2, and the voltage across R is v = R(i 1 i 2 ). i i 1 v R i 2 We will deal extensively with both Nodal Analysis and Mesh Analysis later on. 2

15 Principles of Electrical Engineering I Modeling of Practical Sources All physical devices need to be modeled appropriately in terms of circuit elements so that their characteristics can be taken into account in circuit analysis. In this notes, we emphasize modeling of practical voltage and current sources. Modeling of a practical voltage source: An ideal voltage source keeps the voltage across its terminals the same whatever might be the current (and thus the power) it supplies to a load. Obviously, a practical voltage source can never do so because of its own internal behavior. The voltage across the terminals of a practical voltage source drops more and more as it supplies more and more current. As such, a practical voltage source can be modeled as an ideal voltage source in series with a resistance that can appropriately be called as its internal resistance (in general, impedance). Figure 1 shows a practical voltage source (denoted by α) whose terminals are a and b. A load β when connected to the source draws a current from it. A circuit model of a practical voltage source is shown in Figure 2 along with the load β. Practical Source α i b - v β Load a Figure 1 Model of a Practical Voltage Source i b R g v g v β Load - a Figure 2 i (A) v (V) Table 1 In what follows, we illustrate by an example how to compute v g and R g for a given set of data, and also compute the range of current over which the circuit model of the given practical voltage source is valid. Example 1: A set of experimental data is shown in Table 1 for a practical voltage source. The measurements are obtained experimentally showing how the voltage v across the terminals of a given practical voltage source changes as the current i it supplies to the load varies. Consider the model shown in Figure 2, and determine v g and R g, and the region in which the model is valid.

16 2 Modeling of a practical current source: An ideal current source keeps the current it supplies to a load the same whatever might be the voltage (and thus the power) a load demands. Obviously, a practical current source can never do so because of certain leakages inherent internal to it. The current it supplies to the load decreases as the voltage the load demands increases. As such, a practical current source can be modeled as an ideal current source in parallel with a resistance that can appropriately be called as its internal resistance or leakage resistance. Figure 1 shows a practical current source (denoted by α) whose terminals are a and b. A load β when connected to the source demands a voltage across it. A circuit model of a practical current source is shown in Figure 3 along with the load β. Practical Source α i b - v β Load a Figure 1 i g Model of a Practical Current Source R g Figure 3 i b v β Load - a v (V) i (ma) Table 2 In what follows, we illustrate by an example how to compute i g and R g for a given set of data, and also compute the range of voltage over which the circuit model of the given practical current source is valid. Example 2: A set of experimental data is shown in Table 2 for a practical current source. The measurements are obtained experimentally showing how the current i through the terminals of a given practical current source changes as the voltage v it supplies to the load varies. Consider the model shown in Figure 3, and determine i g and R g, and the region in which the model is valid.

17 Principles of Electrical Engineering I A Transistor Amplifier Circuit v in G R B C Q B E Transistor R C v o v g G i b R B B r p v be E g m v be C Transistor Equivalent r o R C v o The circuit on the left is a simple amplifier circuit containing a device called a transistor (the circled component, denoted by Q) having three terminals B, E, and C. Note: The transistor circuit given here is missing its bias voltages which make it work. You will learn the specifics of this circuit in a later course. For some bias voltages, the so called small signal equivalent circuit of transistor circuit is shown on the right. Here r p is the resistance of the so called plate between B and E, g m is the transconductance of the transistor, and r o is the output resistance associated with the transistor. The ratio of the output voltage v o to the input voltage v in is called the amplifier gain A. Compute A. See the other side for a design example.

18 2 A design example: A pressure-gauge generates a voltage proportional to the pressure exerted on it. However, the magnitude of voltage generated is small and need a very sensitive meter to read it. As such, for proper reading, one needs to amplify the voltage generated by the pressure-gauge. The transistor amplifier circuit shown on the other side is to be used to amplify the voltage close to 200 fold, i.e., the amplifier gain A needs to be as close as possible to 200. For the circuit shown (on the other side) v in is the voltage generated by the pressure-gauge. A design engineer, based on other considerations, decides to use R B = 400Ω and R C = 8KΩ, and then looks for an appropriate transistor. He/she finds in a catalog three different transistors whose parameters are tabulated below. Which of the three transistors is the most appropriate one to be used? Transistor r p in KΩ g m r o in KΩ Hint: Determine the gain A for each transistor

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20 332:221 Principles of Electrical Engineering I Tracing a Circuit Determine v x in the circuit shown. a 5A b 50Ω vx c 20Ω d 200Ω e 0.275v x f Often the practical layout of a circuit is completely different to the one required for easy analysis. It is hard to see what is going on in the circuit given above unless you are well experienced. The circuit given needs to be retraced so that all the interconnections are clear. How do we retrace a circuit? Hint: The circuit branches or components between any two nodes in the given circuit as well as the re-traced circuit must be the same, otherwise the given circuit and the re-traced circuit are not equivalent. The retraced circuit is as shown on the right (justify it). We can now write the following equation easily, [ v x ] 5 = a, c, e b, d, f 5A 50Ω vx 20Ω 200Ω 0.275v x This yields v x = 25 V. Having determined v x, we can find next the current in each resistance easily.

21 Principles of Electrical Engineering I Example 1: KCL-KVL Examples Consider the partially solved circuit on the right with α = 10.5 and β = 5. 3 Determine i 1, i x, αi x, i 2, v d, and v g in the order given. Hints: B To determine i x, write a KVL equation around BGCB. To determine i 1, write a KCL equation at C. αi x is easy to determine once i x is known. To determine i 2, write a KCL equation at A. To determine v d, write a KVL equation around AGBA. To determine v g, write a KVL equation around DGCD. αi x =? 10Ω 16A G 60V A 5Ω i 2 =? 12Ω B - v d =? βi 1 D 9 A 5A 10A Ω i x =? v g =? i 1 =? C The complete solution which satisfies both the branch constraints and the intersection constraints is as shown on the right (justify it). 9A αi x = 21A Ω 10Ω 170V i 1 = 12A A 16A G 60V 5Ω i x = 2A C i 2 = 5A 40V βi 1 12Ω - 5A 10A D

22 2 Cutset: Kirchoff s Current Law(KCL) is normally written at a node where several branches meet. KCL can also be written for what are known as cutsets. A cutset of a circuit is a set of branches which when cut divides the circuit into two parts. Let one side of a cutset be denoted as side N and the other side as Side S. Then, KCL states that the sum of currents going from side N to side S equals the sum of currents going from side S to side N. B Example 2: In the circuit shown ontheright, determine i 2 and i 5 by using node equations or cutset equations. 5A R4 N S 16A G 60V A C R 3 i 2 =? i 5 =? N S 10A R 2 R 5 R 1 D v in G i b R B B r p v be E C i x g m v be R C v o Example 3: Determine i x in the circuit shown on the left. The answer is zero. Justify it. Example 4: Some times one does not need to solve the entire circuit to determine just what is needed. In the circuit on the right, determine the output voltage v o. 1Ω i ba B R 1 C 3iy R 2 6A A 3A v o 3Ω i o R 3 G 1Ω 4i x i x D 5A F R 4 i y E

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24 Principles of Electrical Engineering I Determination of open circuit voltage v oc : We note that there is no current through 10Ω resistance. Hence v x = 0. Thus is v oc is same as v oc. This implies that the dependent source current is zero. It is also easy to see that the 2 A current source simply supplies the 5Ω resistance establishing 10 V across it. The open circuit voltage at the terminals a and b is the same as the voltage across the 5Ω resistance, that is 10 V. Thus v oc = 10 V. 10Ω vx b 10Ω vx b 2A 5Ω 0.7v x 2A 5Ω 0.7v x a a 2A 5Ω 10Ω vx 0.7v x b a Determination of Short Circuit Current i sh : Consider the circuit shown where the terminals a and b are short circuited. Mark the unknown node voltage as v 1 with respect to the ground. Note that v x = v 1. We can then write the relevant KCL equation as 2A 5Ω v 1 v x b 10Ω 0.7v x a i sh v 1 10 v 1 5 = 0.7v x 2 = 0.7v 1 2. By solving the above equation, we get v 1 = 5 V. Thus i sh = v 1 = 0.5 A. 10

25 332:221 Principles of Electrical Engineering I Home-work: try to do the problem yourself before reading the solution given on the next page. Consider the circuit shown. We are interested in determining i a, i b, and i c. Use Kirchoff s current law systematically and repetitively to determine them. Q 2A R 1A 3A L i 1 i 2 M N P 5A G i 3 H J K i4 4A 3i 2 2i 1 C D E F i a 3i 4 2i 3 2i 2 i b A i c B

26 2 Solution: Write the KCL equations at the following nodes in that order to solve for each branch current: Node Q, Node R, Node L, Node M, Node P, Node N, Node G, Node H, Node K, Node J, Node C, Node F, Node E Node B, Node A, and Node D. One of the above equations just acts as a check or verification. The answers are i 1 = 1 A, i 2 = 1 A, i 3 = 3 A, i 4 = 2 A, i a = 2 A, i b = 2 A, and i c = 4 A. In this problem, we just tried to illustrate Kirchoff s current law. We could solve here for the required currents by using only Kirchoff s current law. In a general circuit analysis we need both Kirchoff s current and voltage laws to determine the required variables. We will learn how to use these laws as we progress with the course.

27 332:221 Principles of Electrical Engineering I Home-work: try to do the problem yourself before reading the solution given on the next page. Consider the circuit shown. We are interested in determining v c, v d, v e, v f, v j, and v k. Use Kirchoff s voltage law systematically and repetitively to determine them. 10V 20V 30V 3v 4 A M v 1 v 3 B L v 2 v 4 C K v c v k D J v d v j 2v 4 v e E F H 3v 3 v f 2 2v 2v 1 40V 50V 60V G1 G2 G3 G4 G5 G7 G8 G9 G10 G11 G6

28 2 Solution: Write the KVL equations for the following loops in that order to solve for each branch voltage: loop G1ABG2G1, loop G2BCG3G2, loop G11MLG10G11, loop G10LKG9G10, loop G3CDG4G3, loop G4DEG5G4, loop G9KJG8G9, loop G8JHG7G8, loop G7HFG6G7, and loop G5EFG6G5. The answers are v 1 = 30 V, v 2 = 50 V, v 3 = 110 V, v 4 = 90 V, v c = 240 V, v d = 60 V, v e = 610 V, v f = 280 V, v j = 40 V, and v k = 100 V. In this problem, we just tried to illustrate Kirchoff s voltage law. We could solve here for the required voltages by using only Kirchoff s voltage law. In a general circuit analysis we need both Kirchoff s current and voltage laws to determine the required variables. We will learn how to use these laws as we progress with the course.

29 Tracing Circuits Example-1 Show or argue that the following four circuits are equivalent. Do it yourself first before reading the solution given below. L M N A G E Z 4 v g i g ki x Z 3 B Z 1 C i x Z 2 D B Z 1 C i x Z 2 D ki x Z 3 Z 4 v g i g A G E L M N N M L E G A i g v g Z 4 Z 3 ki x D Z 2 C i x Z 1 B D Z 2 C i x Z 1 B Z 3 ki x i g v g Z 4 E G A N M L Note: Note that how the circuit is drawn is not important although visually one could feel better with one way of drawing the circuit than the other. What is important is what is connected between any two designated nodes. Solution: Although the physical configuration or layout of each circuit is different, all the above circuits have exactly the same nodes, and moreover in between any two given nodes exactly the same branches exist. As such, all the circuits are electrically equivalent.

30 332:221 Principles of Electrical Engineering I FALL 2005 Quiz 1 This design problem is a home-work which will be collected and graded. Consider the circuit shown. By writing a KCL at an appropriate node, determine i s. v α b c d α i s v 5i β β s 20V 7A a 2A f Figure 1 e Determine the voltage across the dependent source in the direction shown. By writing a KVL around an appropriate loop, determine the voltage v α across the element α in the direction shown. By writing a KVL around an appropriate loop, determine the voltage v β across the element β in the direction shown. Fill the following table regarding power consumption or generation. A particular element either consumes or generates power. You must put the appropriate value for a particular element in only one of the unfilled columns whichever is appropriate. Element Power Consumed Power Generated Element α Element β Independent Source Dependent Source

31 Consider the so called bridged T circuit that occurs commonly in applications. Some one partially solved the circuit shown on the right and determined v Ge as 12 V. Determine the following by using appropriately, Ohm s law, KVL, and KCL. At each step, state the law you used. Bridged Tee Example c 24V k a i a 2i x d v Ge G 3.6Ω vba e i e 6Ω vfe Determine the voltage of the controlled source 2i x and then determine the controlling current i x. You may write the KVL for the loop kcdegk. 4Ω f b v mh m h i x 1Ω After knowing i x, determine v mh. Determine v fe by using an appropriate KVL (look for a closed path which by now has only v fe as unknown). Determine i e (Use Ohm s law, note that v fe is known by now). Determine i a by using the KCL at the node h. Proceeding in the same way solve for the current in each branch (mark the current values on the circuit) and verify whatever you have done above is correct. That is KCL equation at each node is satisfied and all three meshes, kcdegk, efhmge, and abfeda, satisfy KVL equations.

32 Principles of Electrical Engineering I This Home-Work problem is collected and graded Determination of voltage of node b with respect to node a: This can be done by a number of ways. At this time, we can use KCL and KVL judiciously in a straight forward way. Step 1: Write a KCL at node e to determine the current in 10Ω resistance in terms of the variable i x (Note that i x is yet unknown). 26V i x 2Ω 3Ω 10Ω 0.5i x b G a Step 2: Write a KVL around the closed-loop GcdefG. Solve it to determine i x. Step 3: Walk along the path afgcdb and algebraically add the voltages encountered to determine the voltage of node b with respect to node a.

33 Principles of Electrical Engineering I Determination of short circuit current i sh : This can be done by a number of ways. At this time, we can use KCL and KVL judiciously in a straight forward way. Step 1: Write a KVL for the closed-loop afgcdba to determine i x. 26V i x 2Ω 3Ω b i sh 10Ω 0.5i x G a Step 2: Write a KCL at node d to determine the current in 3Ω resistance in terms of i x and i sh (Note that the variable i sh is yet unknown). Step 3: Write a KCL at node e to determine the current in 10Ω resistance in terms of i x and i sh (Note that the variable i sh is yet unknown). Step 4: Write a KVL for the closed-loop afedba and determine i sh.

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